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In the image below , How do I resolve the vector M into it's three components x,y,z knowing it's magintude M =$20\sqrt5$ and $A=(4,0,0)$ and $B=(2,4,0)$ enter image description here

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  • $\begingroup$ Do show us your efforts. $\endgroup$ Jan 26 at 16:27

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The componets of vector AB are:

$(m, n, l)=(4-2=2, 0-4=-4, 0-0=0)$

Let coponents of M be $(a, b, c)$; condition for perpendiculaty is:

$a\times m+b \times n+c\times l=0 $

substituting values we get:

$2a-4b=0$

We also have:

$M^2=20^2\times 5= a^2+b^2$

Now find a or b from first equation and substitute in second one and find a and b.

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Use cross product:$$\vec M=c(\vec{B}-\vec{A})\times(\vec L-\vec A)$$ From the magnitude of $\vec M$ you can get the $c$ value, using the fact that $(\vec B-\vec A)\perp(\vec L-\vec A)$.$$20\sqrt 5=c\cdot2\sqrt 5\cdot 4\\c=\frac52$$ Then $$\vec L-\vec A=(0,0,4)\\\vec B-\vec A=(2,4,0)-(4,0,0)=(-2,4,0)$$ I leave the last step to you.

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